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OLEGan [10]
3 years ago
5

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.

50×105 V/m . When the space is filled with dielectric, the electric field is E= 2.40×105 V/m .
What is the charge density on each surface of the dielectric?
Physics
1 answer:
svetlana [45]3 years ago
6 0

Explanation:

The given data is as follows.

     Electric field between plates without dielectric, E_{1} = 3.50 \times 10^{5} V/m

   Electric field between the plates with dielectric, E_{2} = 2.40 \times 10^{5} V/m.

  Permittivity of free space, \epsilon_{o} = 8.85 \times 10^{-12} C^{2}/Nm^{2}

Now, we will determine the charge density as follows.

            \sigma_{i} = \epsilon_{o}(E_{1} - E_{2})

                 = 8.85 \times 10^{-12} \times (3.50 \times 10^{5} - 2.40 \times 10^{5})

                 = 9.735 \times 10^{-7} C/m^{2}

Thus, we can conclude that the charge density on each surface of the dielectric is 9.735 \times 10^{-7} C/m^{2}.

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a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

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Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

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Plugging the values we have.

V_{f} =V_{i} + at

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Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

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So s = distance covered.

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Plugging the values.

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